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n200080 [17]
2 years ago
6

The mass percent of oxygen in pure glucose, c6h12o6 is 53.3 percent. a chemist analyzes a sample of glucose that contains impuri

ties and determines that the mass percent of oxygen is 49.7 percent. which of the follow impurities could account for the low mass percent of oxygen in the sample?
Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0
We can calculate the mass percent of an element by dividing its atomic mass by the mass of the compound and then multiply by 100:
     % by mass of element = (mass of element/mass of compound) x100%
Impurities like n-eicosane with the molecular formula C20H42 could account for the low percent by mass of oxygen in the sample because it has a zero percent oxygen based on its compound formula which indicates that it does not have the element oxygen.
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Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
Grace [21]

Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol

Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

3 0
2 years ago
Only one isotope of this element occurs in nature. one atom of this isotope has a mass of 9.123 ✕ 10-23 g. identify the element
o-na [289]

The atom has only one isotope which means 100 % of same atom is present in nature. The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. Mass of one atom of that isotope is 9.123 ✕ 10⁻²³ g, so mass of one mole of atom that is Avogadro's number of atom is 6.023 X 10²³  X 9.123 X 10⁻²³ g=54.94 g = 55 g (approximate).

So, the atom having atomic mass 55 will be Cesium (Cs). Only one isotope of Cesium is stable in nature.

5 0
2 years ago
Read 2 more answers
Solve without using a calculator. Substance ab2 is 60.0% a by mass. What is the percent a by mass for substance ab?
Orlov [11]

The percent A by mass for substance AB =<u> 75%</u>

<h3>Further explanation</h3>

Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements

Empirical formula is the mole ratio of compounds forming elements.

From Substance AB₂ is 60.0% A by mass.

Let's say that AB₂ mass = 100 gram, then

mass A = 60 gram

mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram

In compound AB:

Total mass = mass A + mass B

Total mass = 60 + 20 grams = 80 grams

Then the percentage of compound A = (60: 80) = 75%

<h3>Learn more</h3>

Grams of KO₂ needed to form O₂

brainly.com/question/2823257

Keywords : percent mass, substance

#LearnwithBrainly

5 0
2 years ago
How many moles of AgNO3 must react to form 0.854 mol Ag?
kipiarov [429]
The answer:
<span>The equation of its dissolution in water is: AgNO3 → Ag + (aq)  +  NO3- (aq)

and     </span>AgNO3 →   Ag + (aq)  +  NO3- (aq)
          1 mol          1mol                1mol 
             ?  --------   0.854mo
so for finding the value, it is sufficients to complute 1 x 0.854 mol =0.854 mol
so,  0.854 mol is required for the reaction to form 0.854 mol of Ag

5 0
2 years ago
Read 2 more answers
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

8 0
2 years ago
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